1. Systems Architecture I (CS 281-001) Lecture 1: Random Access Machines
Jeremy R. Johnson
Wed. Sept. 22, 1999
Sept. 22, 1999
Systems Architecture I

2. Systems Architecture I
September 4, 1997
Introduction
Objective: To develop a simple model of computation that provides insight into how a program executes on a computer.
A Random Access Machine (RAM) is an abstract model of computation that resembles a simple idealized computer. It is equivalent in computational power to a Turing machine (can perform any computation). Despite its simplicity it provides some intuition as to how a program executes on a computer.
Sept. 22, 1999
Systems Architecture I

3. Systems Architecture I
September 4, 1997
Definition of a RAM
Defined by a set of instructions and a model of execution.
A program for a RAM is a sequence of instructions.
A RAM has an infinite memory. Instructions can read and write to memory. Items from memory are loaded into registers, where arithmetic can be performed.
The state of a computation: program counter (to keep track of instruction to execute), registers, and memory.
Sept. 22, 1999
Systems Architecture I

4. A Random Access Machine
Control Unit
Memory 1 2
Program 3
... 4 5 6
AC = accumulator
register
...
Sept. 22, 1999
Systems Architecture I

5. Systems Architecture I
September 4, 1997
Instruction Set
LDA X; Load the AC with the contents of memory address X
LDI X; Load the AC indirectly with the contents of address X
STA X; Store the contents of the AC at memory address X
STI X; Store the contents of the AC indirectly at address X
ADD X; Add the contents of address X to the contents of the AC
SUB X; Subtract the contents of address X from the AC
JMP X; Jump to the instruction labeled X
JMZ X; Jump to the instruction labeled X if the AC contains 0
JMN X; Jump to the instruction labeled X if the contents of the AC ; is negative
HLT ; Halt execution
Sept. 22, 1999
Systems Architecture I

6. Systems Architecture I
September 4, 1997
Sample Program
STOR
; algorithm to detect duplicates in an array A of size n.
for i  1 to n do
if B(A(I))  0
then output A(i);
exit
else B(A(i)) = 1
Sept. 22, 1999
Systems Architecture I

7. Systems Architecture I
September 4, 1997
Sample RAM Program
1. LDI 3; get ith entry from A
2. ADD 4; add offset to compute index j
3. STA 5; store index j
4. LDI 5; get jth entry from B
5. JMZ 9; if entry 0, go to 9
6. LDA 3; if entry 1, get index i
7. STA 2; and store it at 2.
8. HLT ; stop execution
9. LDA 1; get constant 1
10. STI 5; and store it in B
11. LDA 3; get index i
12. SUB 4; subtract limit
13. JMZ 8; if i = limit, stop
14. LDA 3; get index i again
15. ADD 1; increment I
16. STA 3; store new value of I
17. JMP 1; AC
Memory 1
constant 1
answer 2 6
Index i 3 9
Limit of A 4
Index j 5 3 6 4 7 A 2 8 2 9 10 11 B 12 13
Sept. 22, 1999
Systems Architecture I

8. Systems Architecture I
September 4, 1997
Exercises
Modify STOR so that when a computation finishes and the input sequence contained a duplicate integer, we know what that integer was.
Write a RAL program which takes to two input integers at addresses 1 and 2 and multiplies them storing the result at address 4.
Sept. 22, 1999
Systems Architecture I

9. Sample Solution (2nd exc) compute x*y, x,y >= 0
September 4, 1997
Sample Solution (2nd exc) compute x*y, x,y >= 0 AC
1. LDA 1; load x
2. JMZ 10; check if x = 0
3. LDA 4; load partial result
4. ADD 2; add y to partial result
5. STA 4; store partial result
6. LDA 1; load x
7. SUB 3; and decrement
8. STA 1; store decremented x
9. JMP 2; next iteration
10. HLT ;
Memory x
value of x 1 y
Value of y 2 1
Constant 1 3
result 4
The program still works with y < 0; however, if x < 0, it will go into an infinite
loop (x will never = 0). To allow x < 0, first check to see if x is negative with
JMN, and if so we want to increment x rather than decrement it.
Sept. 22, 1999
Systems Architecture I